Inverse Marx generator

[Adrian Jansen]

Depends on the definition of "depends"

"Charge" IS conserved. So if you transfer Q from C1 to C2 >>>

C1*V1 == C2*V2

...Jim Thompson

If you conserve energy, then you must have

C1*V1^2 = C2*V2^2

 

[[email protected]]

Your "point" is, as usual, vagueness and subterfuge... and BS.

For amusement, newbies are invited to MATHEMATICALLY analyze this
simple case...

Two equal value capacitors (C), one charged to Vo, the other uncharged
(zero volts).

Connect together with a switch, start with a finite resistance value,
analyze; then reduce the resistance, re-analyze; continue this
analysis, approaching zero in the limit.

Then scratch your head in surprise... where did the energy go ?

Where it always goes.

 

[JosephKK]

If you conserve energy, then you must have

C1*V1^2 = C2*V2^2

By analogy, consider a billiard ball striking another (at rest) ball 1/4
diameter away from the line of centroid motion (glancing blow / cut).
Does energy and momentum get conserved? Show the math.

 

[Graeme Zimmer]

Thanks Fred,

I would have thought that as the Capacitor closes (eg C increases ) the
Voltage decreases, but net Charge remains the same..

Sort of like going from High Impedance to Low Impedance.
It's trading Voltage for Current, which is what we want.

I agree the attraction between the plates will drive the Cap like a motor.
Is this good or bad?


...................... Zim

 

[Chieftain of the Carpet Crawlers]

I CAN do the math! Can others here? Why do you think I was dumping
on Larkin?

First Law of Thermodynamics: You always lose

...Jim Thompson

A pair of billiard balls colliding, even at perfect tangency, does not
do so without losses.

In the glancing blow, the coefficient of friction of the ball media
comes into play as does the coefficient of friction of the ball-to-cloth
interface.

These two factors mean that there will not be a 100% transfer of energy
from one ball to the other without some loss.

Otherwise, 'English' would not work. Nor would 'Throw'. Two very
important weapons in the billiard artist's arsenal. Bank shots would
also have several issues.

Also, using a 'term' like "a quarter ball" can be ambiguously
interpreted.

Just give the number of degrees in the future.

Question/poser for you all:

Can a 'cut shot' be made that is less than 90° with these friction
effects in place?

The answer is yes, but do you know how or why to get there?

 

[Fred Abse]

Thanks Fred,

I would have thought that as the Capacitor closes (eg C increases ) the
Voltage decreases, but net Charge remains the same..

Yes, that's what would happen.

However, the same charge at a lower voltage represents less energy. The
energy lost has to go somewhere. In this case it will go into the rotating
system.

Take a simpler example: two parallel plates moved closer, or separated.
Separation will require work to be done on the plates to overcome
attraction. Moving closer, the plates will do work themselves. Energy will
be conserved.

Sort of like going from High Impedance to Low Impedance. It's trading
Voltage for Current, which is what we want.

The concept of impedance doesn't really apply here.

I agree the attraction between the plates will drive the Cap like a
motor. Is this good or bad?

It's bad, since the object of the exercise is to convert energy at a high
voltage to energy at a lower voltage. We want to finish with the joules
that we started with, neglecting losses.

 

[Fred Abse]

Jim Thompson wrote:
If you conserve energy, then you must have

C1*V1^2 = C2*V2^2

Both conditions are only satisfied when C1=C2

 

[Winfield Hill]

Jim Thompson wrote...

Jim Thompson wrote: [snip]

Depends on the definition of "depends"
"Charge" IS conserved. So if you transfer Q from C1 to C2 >>>
If you conserve energy, then you must have
C1*V1^2 = C2*V2^2

Right. If you dump all the energy from one charged cap into
another, discharged, cap of a different value, and do it
efficiently, charge is not conserved.

John says, "...charge is not conserved."
Newbies are invited to Google on "conservation of charge".
(AND run the math problem I previously posted.)
John is so full of it I'd bet his eyes are brown ;-)

Unfortunately, Adrian Jansen mis-states the results as well :-(

I haven't been following this thread, but I have a comment.

The operative phrase must be, "and do it efficiently."

This is easy to do, with a dc-dc converter for example, or a
mosfet switch and an inductor. In these cases it's easy to
manipulate E1 and E2, C1*V1^2 = C2*V2^2. Forget about charge.

 

[Chieftain of the Carpet Crawlers]

Billiard balls roll, which means that at the instant of contact there
is a transverse scraping, like a clutch engaging, which wastes more
energy than a classic elastic sphere conservation-of-momentum physics
problem. The angular momentum transfer is a lot like transferring
charge between two capacitors by connecting them with a resistor, but
worse because of the felt.

Air hockey is closer to classic elastic collision.

John

It appears that we agree on something today.

The "clutching" you refer to is due to the inertia of the spin on the
rolling ball imparting a torqueing force to the ball it makes contact
with in the fraction of a second that the "less than 100% elastic
collision" takes to occur, and is, of course 100% related to the
coefficient of friction between the two colliding masses.

"Air Hockey" is actually about as close as one can get to modeling the
ideal circumstance of a pure elastic collision with zero losses here on
this globe's surface (table MUST be 100% level). Ill bet that it would
be cool to experiment with different "puck" materials. The puck must
also have very, VERY 'true' side faces. That is to say... they must be
square to the puck faces.

I am an advanced Masse' 'applicator' in billiards. I can even compress
a ball against a rail nose so hard that it compresses it shape into it
before rebounding. This allows for a lot more manipulation of the energy
one places into the rubber rail.

If I use "draw" on the cue ball, it imparts "follow" on the object
ball, which then causes it to 'loft' a bit upon compression of the rail.
This makes the rebound event occur at a plane other than that of the
table bed (i.e. 1,2, 3, or 4 degrees a.o.a.)

I can bank a ball "cross-side" 100% in the air, after the rail contact
is made, if I 'loft' it in such a manner. I can do it full table length
as well, but it takes a lot more 'effect' imparted to it to perform. One
actually has to bear down on the ball such that it takes little hops on
its way down to the end rail. Otherwise a lot of the applied English
gets removed by the cloth and the shot fails. It is usually about two or
three little bounces that psuedo-float the ball down the table.

 

[AM]

We've gone from "dumping" to running a switcher.

Got one at 100%? Power-Out == Power-In ??

...Jim Thompson

So, is the question then:

Does transferring charge have a cost? Is "work" required to perform
the task?

 

[Winfield Hill]

Jim Thompson wrote...

Jim Thompson wrote...

John Larkin wrote:
Adrian Jansen wrote:
Jim Thompson wrote:
[snip]

Depends on the definition of "depends"
"Charge" IS conserved. So if you transfer Q from C1 to C2 >>>

If you conserve energy, then you must have
C1*V1^2 = C2*V2^2

Right. If you dump all the energy from one charged cap into
another, discharged, cap of a different value, and do it
efficiently, charge is not conserved.

John says, "...charge is not conserved."
Newbies are invited to Google on "conservation of charge".
(AND run the math problem I previously posted.)
John is so full of it I'd bet his eyes are brown ;-)

Unfortunately, Adrian Jansen mis-states the results as well :-(

I haven't been following this thread, but I have a comment.

The operative phrase must be, "and do it efficiently."

This is easy to do, with a dc-dc converter for example, or a
mosfet switch and an inductor. In these cases it's easy to
manipulate E1 and E2, C1*V1^2 = C2*V2^2. Forget about charge.

"Forget about charge."???
I guess it depends on your definition of "depends" :-(
We've gone from "dumping" to running a switcher.

I imagine one can use the word dump. Consider this setup:
A charged C1 with switch S1 to the top of grounded inductor
L1, and from the inductor another switch S2 to a discharged
cap C2. Close switch S1 and the inductor current builds,
open S1 when C1 is discharged to 0V and the L1 current has
reached a peak. Simultaneously close S2 and L1's current
continues to flow, charging C2, until the inductor current
stops, then open S2. (One can avoid time-precise switches
by using series diodes, but they do have loss.)

So C1's energy is dumped first into L1, and then into C2.

If C2 is much smaller than C1, it'll end up with a much
higher voltage, and the charge out of C1 will greatly
exceed the charge into C2 (same current, shorter time),
even though E1 = E2, exactly, at least in principle.
It must meet John's definition of "do it efficiently."

Got one at 100%? Power-Out == Power-In ??

I made one of these where C1's starting voltage was 1.2kV,
and the inductor's peak current was about 1.5kA. I used
large IGBT switch modules with Vce(sat) of 3V, or about
0.3% loss. Pulse cap esr and a litz-wire inductor were
responsible for most of the losses, about 5% or so, IIRC.

 

[AM]

On the other hand, energy is always conserved.

John

If you are around to observe and tell about it.

 

[AM]

If you are around to observe and tell about it.

B A N G !!! Energy finds its way out of the box(es)!

 

[Perenis]

Well, Heaven help us, Larkin could "slosh" forever

Like the 'pile' in the Oxford bell jar?

The 'clapper' actually.

 

[AM]

Cite?

The length of time that you'll deny it!

 

[AM]

It helps to understand ideal circuits before you consider real
circuits. The ideals are the limiting cases. You CAN transfer charge
between equal value caps without loss of charge, and you can more
generally transfer energy between caps without loss; just use an
inductor.

Why then does it not work with a full and empty battery?

What is the final voltage the battery pair will be at after
the same such "transfer".

Also, in both cases, how do you propose to do it without inrush damage
to the empty cap/battery and outflow damage to the full cap/battery?

There are losses, because the cap has terminations that are not ideal
in nature. There will also be damage sites along the contact face for
the cap plate-to-terminal_interface interface.

You cannot do it with ideal caps because the charge current would jump
to infinity.

There MUST be resistance in the circuit to limit the charge current.

 

[AM]

Some of TI's buck-boost configurations look a bit like what you
describe:

http://focus.ti.com/lit/ds/symlink/tps63020.pdf

kevin

I wonder if they could make a MEMS "Oxford Electric Bell" and use it as
a precision clock or counter even.

 

[JosephKK]

Billiard balls roll, which means that at the instant of contact there
is a transverse scraping, like a clutch engaging, which wastes more
energy than a classic elastic sphere conservation-of-momentum physics
problem. The angular momentum transfer is a lot like transferring
charge between two capacitors by connecting them with a resistor, but
worse because of the felt.

Air hockey is closer to classic elastic collision.

John

If you like air hockey pucks better do it for them instead. You can even
assume any spins you want including no spins. Please continue.

 

[JosephKK]

Jim Thompson wrote...

John Larkin wrote:
Adrian Jansen wrote:
Jim Thompson wrote:
[snip]

Depends on the definition of "depends"
"Charge" IS conserved. So if you transfer Q from C1 to C2 >>>

If you conserve energy, then you must have
C1*V1^2 = C2*V2^2

Right. If you dump all the energy from one charged cap into
another, discharged, cap of a different value, and do it
efficiently, charge is not conserved.

John says, "...charge is not conserved."
Newbies are invited to Google on "conservation of charge".
(AND run the math problem I previously posted.)
John is so full of it I'd bet his eyes are brown ;-)

Unfortunately, Adrian Jansen mis-states the results as well :-(

I haven't been following this thread, but I have a comment.

The operative phrase must be, "and do it efficiently."

This is easy to do, with a dc-dc converter for example, or a
mosfet switch and an inductor. In these cases it's easy to
manipulate E1 and E2, C1*V1^2 = C2*V2^2. Forget about charge.

Exactly. To say "Charge is always conserved" is absurd. It is
conserved in some situations, not in others. The context must be
stated exactly.

Charge two identical caps to the same voltage, then connect them in
parallel, but with polarities flipped. ALL the charge vanishes.

On the other hand, energy is always conserved.

John

Well let's consider this test case you just described. There was energy
stored in each capacitor before closing the switch. There is none
afterwards. Where did it go? How did it get there?

 

[The Great Attractor]

If you like air hockey pucks better do it for them instead. You can even
assume any spins you want including no spins. Please continue.

So. Mr. Guru, is the Universe going to collapse back in,back to the
point at which it began, or expand forever? Or will it go out with a
blink and just disintegrate?